The chain rule

When you need it

• The chain rule is for a function inside another function — a composite function.

• $(3x + 1)^5$ is "raise to the fifth power" applied to "$3x+1$".

• $\sin(x^2)$ is "take the sine" applied to "$x^2$".

• $e^{\cos x}$ is "exponentiate" applied to "$\cos x$".

• Recognising a composite is the skill. Ask: if I had to evaluate this for $x = 2$, what would I compute first? That inner computation is the inside function.

The rule

$$\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}$$

• Or, in the form most people use:

$$\frac{d}{dx}\left[f(g(x))\right] = f'(g(x)) \times g'(x)$$

• In words: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
• The last factor is what students call "the multiplier", and dropping it is the single most-reported error in this standard.

The substitution method

• The safest way to set out a chain rule question:

  • Let $u$ be the inside function.
  • Write $y$ in terms of $u$.
  • Differentiate both: find $\dfrac{du}{dx}$ and $\dfrac{dy}{du}$.
  • Multiply them together.
  • Substitute the inside function back in, so the answer is in terms of $x$.

• Worked through for $y = (3x + 1)^5$:

  • Let $u = 3x + 1$, so $y = u^5$
  • $\dfrac{du}{dx} = 3$ and $\dfrac{dy}{du} = 5u^4$
  • $$\dfrac{dy}{dx} = 5u^4 \times 3 = 15u^4 = 15(3x+1)^4$$

The standard patterns

• Once the method is fluent, these can be written straight down:

$$\frac{d}{dx}\left[(ax+b)^n\right] = an(ax+b)^{n-1}$$

$$\frac{d}{dx}\left[\sin u\right] = u'\cos u \qquad \frac{d}{dx}\left[\cos u\right] = -u'\sin u$$

$$\frac{d}{dx}\left[e^u\right] = u'e^u \qquad \frac{d}{dx}\left[\ln u\right] = \frac{u'}{u}$$

Chains inside chains

• Some functions need the rule applied twice, working from the outside in.

• Worked through for $y = \sin^3(2x)$, which is $\left(\sin(2x)\right)^3$:

  • Outermost: cube, so $3\left(\sin 2x\right)^2$
  • Next layer: derivative of $\sin 2x$ is $2\cos 2x$
  • Multiply: $\dfrac{dy}{dx} = 3\sin^2(2x) \times 2\cos 2x = 6\sin^2(2x)\cos 2x$

• Note that each layer contributes a factor. A three-layer function has three factors in its derivative.

Roots as powers

• Always convert before applying the rule:

$$y = \sqrt{x^2 + 4} = \left(x^2+4\right)^{1/2}$$

$$\frac{dy}{dx} = \frac{1}{2}\left(x^2+4\right)^{-1/2} \times 2x = \frac{x}{\sqrt{x^2+4}}$$

Fractions that are really chains

• $\dfrac{1}{(2x-5)^3}$ looks like a job for the quotient rule, but writing it as $(2x-5)^{-3}$ makes it a chain rule problem, which is much shorter:

$$\frac{d}{dx}\left[(2x-5)^{-3}\right] = -3(2x-5)^{-4} \times 2 = \frac{-6}{(2x-5)^4}$$

• Whenever the numerator is a constant, use a negative power and the chain rule. Reaching for the quotient rule there is extra work for extra risk.